如何在lambda表达式中重新抛出异常？[副本]

Question

如何在Lambda中处理和重新抛出异常？当我试图用try/catch块包围调用时，它只是在lambda表达式中捕获它。我在方法签名中有ServiceException，所以我只想重新抛出它。是可能还是我漏掉了什么？

 BiFunction<Account, LocalDate, AccountBalance> getBalance;

 getBalance = (account, date1) -> balanceProvider.getBalance(account, date1); //balanceProvider.getBalance unhandled ServiceException

Answer 1

我过去常常捕获检查过的异常并重新抛出未检查的异常。类似于：

IntStream.range(1,3).forEach(i -> {
        try {
            TimeUnit.SECONDS.sleep(10);
        } catch (InterruptedException e) {
            throw new RuntimeException("InterruptedException was caught");
        }
    });

或者在你的情况下：

getBalance = (account, date1) -> {
    try {
        balanceProvider.getBalance(account, date1)
    } catch (ServiceException e) {
        throw new ServiceRTException("blabla");
    }
};